Characteristic Equations

This is a second-order linear DE with constant coefficients. We can try \( y = e^{kt}\) and solve the characteristic equation:

\( k^2 - 16k + 48 = 0 \)

\( b^2 - 4ac = (-16)^2 - 4(1)(48) = 64 > 0 \)

The determinant is positive meaning there are two real roots. Factoring gives:

\( (k-4)(k-12) = 0 \)

The general solution is \( y = c_1 e^{4t} + c_2 e^{12t} \).

Again, this is a second-order linear DE with constant coefficients. Calcualte the determinant:

\( b^2 - 4ac = (10)^2 - 4(1)(25) = 0 \)

The determinant is zero meaning there are repeated roots. Factoring gives:

\( (k-5)(k-5) = 0 \)

The general solution is \( y = c_1 e^{5t} + c_2 te^{5t} \).

Note that the second term is actually found using the reduction of order technique.

Again, this is a second-order linear DE with constant coefficients. Calcualte the determinant:

\( b^2 - 4ac = (0)^2 - 4(1)(25) = -100 < 0 \)

The determinant is negative meaning there are complex roots. Using the quadratic equation gives:

\( \cfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \cfrac{-0 \pm (10i)}{2} = 0 \pm 5i \)

The general solution is \( y = c_1 \cos {5 t} + c_2 \sin {5 t} \) since \(\alpha = 0\).

Note that this technique expands to higher-order homogeneous differential equations with constant coefficients. Suppose the characteristic equation is:

\( (k+2)^2(k-1)(k^2-2k+3) \)

The first term has repeated roots giving solution \(c_1e^{-2t} + c_2 t e^{-2t}\).

The second term has a real root giving solution \(c_3e^t\).

The third term has complex roots in the form \( 1 \pm \sqrt{2} \) giving solution \( e^{ t}(c_4 \cos {\sqrt{2} t} + c_5 \sin {\sqrt{2} t}) \)

Putting it all together gives the general solution to the DE:

\( y = c_1e^{-2t} + c_2 t e^{-2t} + c_3e^t + e^{ t}(c_4 \cos {\sqrt{2} t} + c_5 \sin {\sqrt{2} t})\).